Range of a^4+b^4+c^4

81/27

HINT:
a⁴+b⁴+c⁴=p
{a²+b²}²-2(a²b²)=p-c⁴
(2-c²)²-2[(1-c²)-(a²+b²)2]=p-c⁴
Then p=[(4-4c²+c⁴)-(1-c²)-(2-c²)+c⁴]
Now the expr. is a function of c...and can be further factorised....
Hope there are no silly mistakes.... if so....please forgive me.....
Sorry in advance

The square of a^2+b^2+c^2 is a^4+b^4+c^4+2(a^2*b^2+a^2*c^2+b^2*c^2)=4, therefore
a^4+b^4+c^4=4-2(a^2*b^2+a^2*c^2+b^2*c^2)
Using AM-GM, (a^2*b^2+a^2*c^2+b^2*c^2)/3 is >or=to (the cube root of a^2*b^2*a^2*c^2*b^2*c^2),
a^2*b^2+a^2*c^2+b^2*c^2 is >or=to 3*(the cube root of a^4*b^4*c^4), so the maximum value is
4-6*(the cube root of a^4*b^4*c^4)
using the AM-GM on a^4+b^4+c^4, you get a^4+b^4+c^4>or=to 3*(the cube root of a^4*b^4*c^4), the minimum value.

We have using the given conditions

a^4+b^4+c^4 = 4abc + \frac{7}{2}

So we need to find the range of abc

Now we also get ab+bc+ca = -\frac{1}{2}

So a,b,c are roots of x^3-x^2-\frac{x}{2}+t=0

and t =-abc

Now the numbers that come are not nice, so i will give an outline:

1. find the max and min of the polynomial

2. impose the condition that max>0 and min <0 to get bounds for t and hence for abc

3. now find the max and min for given expression

I did a calculation and obtained (2.19, 2.93)

hope i am right :D